Question

我正在尝试编写Python代码，以查看平均需要多少掷硬币才能连续获得N个头部。

我不解的是，我的代码产生的答案与在线给出的答案不匹配，例如在这里（以及其他许多地方）https://math.stackexchange.com/questions/364038/expected-number-of-coin-tosses-to-get-five-consecutive-heads

据此，我应该连续获得各种数量的头部所需的预期投掷次数为：E（1）= 2，E（2）= 6，E（3）= 14，E（ 4）= 30，E（5）=62。但是我没有得到这些答案！例如，我得到E（3）= 8，而不是14。下面的代码给出了答案，但是您可以更改n来测试行中其他目标头的数量。

出了什么问题？我的代码逻辑上大概有一些错误，但是我承认我无法弄清楚它是什么。

您可以在这里查看，运行和修改我的代码：https://trinket.io/python/17154b2cbd

下面是该代码本身，位于该可运行的trinket.io页面之外。任何帮助找出问题所在将不胜感激！

非常感谢，

拉杰附言我能找到的最接近的相关问题是这个：Monte-Carlo Simulation of expected tosses for two consecutive heads in python 但是，据我所知，该问题中的代码实际上并没有测试两个连续头，而是测试了一个以头开始的序列，然后再测试某个以后的序列，可能不是连续的，时间又来了。

# Click here to run and/or modify this code:
# https://trinket.io/python/17154b2cbd

import random
# n is  the target number of heads in a row
# Change the value of n, for different target heads-sequences
n = 3  
possible_tosses = [ 'h', 't' ]
num_trials = 1000
target_seq = ['h' for i in range(0,n)]
toss_sequence = []
seq_lengths_rec = []

for trial_num in range(0,num_trials):

    if (trial_num % 100) == 0:
        print 'Trial num', trial_num, 'out of', num_trials
        # (The free version of trinket.io uses Python2)

    target_reached = 0
    toss_num = 0

    while target_reached == 0:

        toss_num += 1
        random.shuffle(possible_tosses)
        this_toss = possible_tosses[0]
        #print([toss_num, this_toss])
        toss_sequence.append(this_toss)
        last_n_tosses = toss_sequence[-n:]
        #print(last_n_tosses)
    if last_n_tosses == target_seq:
        #print('Reached target at toss', toss_num)
        target_reached = 1
        seq_lengths_rec.append(toss_num)

print 'Average', sum(seq_lengths_rec) / len(seq_lengths_rec)

Answer 1

您不会为每个实验重新初始化toss_sequence，因此您将以预先存在的头部序列开始每个实验，每次尝试时都有1分之二的机会击中目标序列新实验。

在外部循环中初始化toss_sequence将解决您的问题：

import random
# n is  the target number of heads in a row
# Change the value of n, for different target heads-sequences
n = 4
possible_tosses = [ 'h', 't' ]
num_trials = 1000
target_seq = ['h' for i in range(0,n)]
seq_lengths_rec = []

for trial_num in range(0,num_trials):

    if (trial_num % 100) == 0:
        print('Trial num {} out of {}'.format(trial_num, num_trials))
        # (The free version of trinket.io uses Python2)

    target_reached = 0
    toss_num = 0
    toss_sequence = []

    while target_reached == 0:

        toss_num += 1
        random.shuffle(possible_tosses)
        this_toss = possible_tosses[0]
        #print([toss_num, this_toss])
        toss_sequence.append(this_toss)
        last_n_tosses = toss_sequence[-n:]
        #print(last_n_tosses)
        if last_n_tosses == target_seq:
            #print('Reached target at toss', toss_num)
            target_reached = 1
            seq_lengths_rec.append(toss_num)

print(sum(seq_lengths_rec) / len(seq_lengths_rec))

您可以稍微简化代码，并减少出错的可能性：

import random
# n is  the target number of heads in a row
# Change the value of n, for different target heads-sequences
n = 3
possible_tosses = [ 'h', 't' ]
num_trials = 1000
seq_lengths_rec = []

for trial_num in range(0, num_trials):

    if (trial_num % 100) == 0:
        print('Trial num {} out of {}'.format(trial_num, num_trials))
        # (The free version of trinket.io uses Python2)

    heads_counter = 0
    toss_counter = 0

    while heads_counter < n:
        toss_counter += 1

        this_toss = random.choice(possible_tosses)

        if this_toss == 'h':
            heads_counter += 1
        else:
            heads_counter = 0
    seq_lengths_rec.append(toss_counter)


print(sum(seq_lengths_rec) / len(seq_lengths_rec))

Answer 2

我们可以通过将每个实验运行足够长的时间（理想情况下是无限次）来消除一个额外的循环，例如，每次抛硬币 n=1000 次。现在，每次这样的试验中很可能会出现 5 个正面的序列。如果确实出现，我们可以称该试验为有效试验，否则我们可以拒绝该试验。

最后，我们可以取所需的抛掷次数的平均值 w.r.t.有效试验的次数（根据 LLN，它将近似于预期的投掷次数）。考虑以下代码：

N = 100000 # total number of trials
n = 1000 # long enough sequence of tosses
k = 5 # k heads in a row
ntosses = []
pat = ''.join(['1']*k)
effective_trials = 0
for i in range(N): # num of trials
    seq = ''.join(map(str,random.choices(range(2),k=n))) # toss a coin n times (long enough times)
    if pat in seq:
        ntosses.append(seq.index(pat) + k)
        effective_trials += 1
print(effective_trials, sum(ntosses) / effective_trials)
# 100000 62.19919

请注意，如果 n 很小，结果可能不正确，因为它试图近似无限次抛硬币（为了找到连续获得 5 次正面的预期抛硬币次数，n=1000 是可以的，因为实际预期值是 62）。

问：在Python中，预期的抛硬币次数将连续获得N个头。我的代码给出的答案与已发布的正确答案不匹配，但不确定为什么

2 个答案: